Question

In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.​

Answer 1

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

⠀⠀⠀⠀⠀⠀⠀⠀⠀ [ attachment no. (1) ]

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

We know,

Area of trapezium$= \frac{1}{2} × sum\:of\:parallel\:sides × height.$

So, height has to be found.

In the diagram, draw CL || AD

⠀⠀⠀⠀⠀[diagram (2) in the attachment ]

Now, ALCD is a parallelogram $⇒ AL = CD = 20$ cm and $CL = AD = 26\:cm$

As AD = CB,

$CL = CB ⇒ ΔCLB$ is an isosceles triangle with CB as its height.

Here, $BL = AB – AL = (40 – 20) = 20$cm. So,

$LM = MB = \frac{1}{2} BL = \frac{1}{2} × 20 = 10\: cm$

Now, in ΔCLM,

$CL² = CM² + LM²$ (Pythagoras Theorem)

$26² = CM² + 10²$

$CM² = 26² – 10²$

Using algebraic identities, we get; $26² – 10² = (26 – 10) (26 + 10)$

hence,

$CM²= (26 – 10) (26 + 10) = 16 × 36 = 576$

$CM = \sqrt{576} = 24\:cm$

Now, the area of trapezium can be calculated.

Area of trapezium,$ABCD = \frac{1}{2} × (AB + CD) × CM$

$=\frac{1}{2} × (20 + 40) × 24$

Or, Area of trapezium ABCD = 720 cm²